The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants inan agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = 10,000 mAssuming that sucrose is produced by the reactionm²).12CO2 (9) + 11H2O(1)→ C12H22O11(s) + 120O2(g) AH=5640 kJcalculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis inthis field.Percent efficiency

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The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = Assuming that sucrose is produced by the reaction %3D = 10,000 m²). 12CO2 (9) + 11H20(1) → C12H2»O11(s) + 1202(g) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis in this field. Percent efficiency =

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = Assuming that sucrose is produced by the reaction %3D = 10,000 m²). 12CO2 (9) + 11H20(1) → C12H2»O11(s) + 1202(g) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis in this field. Percent efficiency =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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